Question

Taking Bohr's radius as \(r_0=53\,\text{pm}\), the ground state radius of \(Li^{2+}\) ion, on the basis of Bohr's model, will be about

• A. \(53\,\text{pm}\)
• B. \(27\,\text{pm}\)
• C. \(18\,\text{pm}\)
• D. \(13\,\text{pm}\)

Hint:

For hydrogen-like species: \[ r_n=\frac{n^2r_0}{Z} \] Radius is inversely proportional to atomic number \(Z\). Therefore, \(Li^{2+}\) has a radius one-third that of hydrogen in the ground state.

Answer 1

The Correct Option is C

Concept: For a hydrogen-like atom, Bohr's radius of the \(n^{\text{th}}\) orbit is \[ r_n=\frac{n^2r_0}{Z} \] where \[ r_0=53\,\text{pm} \] is the Bohr radius of hydrogen and \(Z\) is the atomic number.

Step 1:Write the values for \(Li^{2+}\). For \(Li^{2+}\), \[ Z=3 \] Ground state corresponds to \[ n=1 \]

Step 2: Calculate the radius. \[ r=\frac{n^2r_0}{Z} \] \[ r=\frac{(1)^2(53)}{3} \] \[ r=\frac{53}{3} \] \[ r=17.67\,\text{pm} \] \[ r\approx18\,\text{pm} \]

Step 3: State the answer. \[ { r_{Li^{2+}}\approx18\,\text{pm} } \] Hence, the correct option is \[ {(C)} \]